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5.3E: Nonhomgeneous Linear Equations (Exercises)

• • Contributed by William F. Trench
• Andrew G. Cowles Distinguished Professor Emeritus (Mathamatics) at Trinity University

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In Exercises [exer:5.3.1} –[exer:5.3.6} find a particular solution by the method used in Example 5.3.2}. Then find the general solution and, where indicated, solve the initial value problem and graph the solution.

[exer:5.3.1] $$y''+5y'-6y=22+18x-18x^2$$

[exer:5.3.2] $$y''-4y'+5y=1+5x$$

[exer:5.3.3] $$y''+8y'+7y=-8-x+24x^2+7x^3$$

[exer:5.3.4] $$y''-4y'+4y=2+8x-4x^2$$

[exer:5.3.5] $$y''+2y'+10y=4+26x+6x^2+10x^3, \quad y(0)=2, \quad y'(0)=9$$

[exer:5.3.6] $$y''+6y'+10y=22+20x, \quad y(0)=2,\; y'(0)=-2$$

[exer:5.3.7] Show that the method used in Example

Example $$\PageIndex{1}$$:

Add text here. For the automatic number to work, you need to add the “AutoNum” template (preferably at5.3.2} will not yield a particular solution of

$y''+y'=1+2x+x^2; \eqno{\rm (A)}$

that is, (A) does’nt have a particular solution of the form $$y_p=A+Bx+Cx^2$$, where $$A$$, $$B$$, and $$C$$ are constants.
 [exer:5.3.8] $$x^2y''+7xy'+8y={6\over x}$$ [exer:5.3.9] $$x^2y''-7xy'+7y=13x^{1/2}$$
 [exer:5.3.10] $$x^2y''-xy'+y=2x^3$$ [exer:5.3.11] $$x^2y''+5xy'+4y={1\over x^3}$$
 [exer:5.3.12] $$x^2y''+xy'+y=10x^{1/3}$$ [exer:5.3.13] $$x^2y''-3xy'+13y=2x^4$$

[exer:5.3.14] Show that the method suggested for finding a particular solution in Exercises [exer:5.3.8} -[exer:5.3.13} will not yield a particular solution of

$x^2y''+3xy'-3y={1\over x^3}; \eqno{\rm (A)}$

that is, (A) doesn’t have a particular solution of the form $$y_p=A/x^3$$.

[exer:5.3.15] Prove: If $$a$$, $$b$$, $$c$$, $$\alpha$$, and $$M$$ are constants and $$M\ne0$$ then

$ax^2y''+bxy'+cy=M x^\alpha$

has a particular solution $$y_p=Ax^\alpha$$ ($$A=$$ constant) if and only if $$a\alpha(\alpha-1)+b\alpha+c\ne0$$.
 [exer:5.3.16] $$y''+5y'-6y=6e^{3x}$$ [exer:5.3.17] $$y''-4y'+5y=e^{2x}$$

[exer:5.3.18] $$y''+8y'+7y=10e^{-2x}, \quad y(0)=-2,\; y'(0)=10$$

[exer:5.3.19] $$y''-4y'+4y=e^{x}, \quad y(0)=2,\quad y'(0)=0$$

 [exer:5.3.20] $$y''+2y'+10y=e^{x/2}$$ [exer:5.3.21] $$y''+6y'+10y=e^{-3x}$$

[exer:5.3.22] Show that the method suggested for finding a particular solution in Exercises [exer:5.3.16} -[exer:5.3.21} will not yield a particular solution of

$y''-7y'+12y=5e^{4x}; \eqno{\rm (A)}$

that is, (A) doesn’t have a particular solution of the form $$y_p=Ae^{4x}$$.

[exer:5.3.23] Prove: If $$\alpha$$ and $$M$$ are constants and $$M\ne0$$ then constant coefficient equation

$ay''+by'+cy=M e^{\alpha x}$

has a particular solution $$y_p=Ae^{\alpha x}$$ ($$A=$$ constant) if and only if $$e^{\alpha x}$$ isn’t a solution of the complementary equation.

[exer:5.3.24] $$y''-8y'+16y=23\cos x-7\sin x$$

[exer:5.3.25] $$y''+y'=-8\cos2x+6\sin2x$$

[exer:5.3.26] $$y''-2y'+3y=-6\cos3x+6\sin3x$$

[exer:5.3.27]$$y''+6y'+13y=18\cos x+6\sin x$$

[exer:5.3.28] $$y''+7y'+12y=-2\cos2x+36\sin2x, \quad y(0)=-3,\quad y'(0)=3$$

[exer:5.3.29] $$y''-6y'+9y=18\cos3x+18\sin3x, \quad y(0)=2,\quad y'(0)=2$$

[exer:5.3.30] Find the general solution of

$y''+\omega_0^2y =M\cos\omega x+N\sin\omega x,$

where $$M$$ and $$N$$ are constants and $$\omega$$ and $$\omega_0$$ are distinct positive numbers.

[exer:5.3.31] Show that the method suggested for finding a particular solution in Exercises [exer:5.3.24} -[exer:5.3.29} will not yield a particular solution of

$y''+y=\cos x+\sin x; \eqno{\rm (A)}$

that is, (A) does not have a particular solution of the form $$y_p=A\cos x+B\sin x$$.

[exer:5.3.32] Prove: If $$M$$, $$N$$ are constants (not both zero) and $$\omega>0$$, the constant coefficient equation

$ay''+by'+cy=M\cos\omega x+N\sin\omega x \eqno{\rm (A)}$

has a particular solution that’s a linear combination of $$\cos\omega x$$ and $$\sin\omega x$$ if and only if the left side of (A) is not of the form $$a(y''+\omega^2y)$$, so that $$\cos\omega x$$ and $$\sin\omega x$$ are solutions of the complementary equation.

[exer:5.3.33] $$y''+5y'-6y=22+18x-18x^2+6e^{3x}$$ (See Exercises [exer:5.3.1} and [exer:5.3.16}.)

[exer:5.3.34] $$y''-4y'+5y=1+5x+e^{2x}$$ (See Exercises [exer:5.3.2} and [exer:5.3.17}.)

[exer:5.3.35] $$y''+8y'+7y=-8-x+24x^2+7x^3+10e^{-2x}$$ (See Exercises [exer:5.3.3} and [exer:5.3.18}.)

[exer:5.3.36] $$y''-4y'+4y=2+8x-4x^2+e^{x}$$ (See Exercises [exer:5.3.4} and [exer:5.3.19}.)

[exer:5.3.37] $$y''+2y'+10y=4+26x+6x^2+10x^3+e^{x/2}$$ (See Exercises [exer:5.3.5} and [exer:5.3.20}.)

[exer:5.3.38] $$y''+6y'+10y=22+20x+e^{-3x}$$ (See Exercises [exer:5.3.6} and [exer:5.3.21}.)

[exer:5.3.39] Prove: If $$y_{p_1}$$ is a particular solution of

$P_0(x)y''+P_1(x)y'+P_2(x)y=F_1(x)$

on $$(a,b)$$ and $$y_{p_2}$$ is a particular solution of

$P_0(x)y''+P_1(x)y'+P_2(x)y=F_2(x)$

on $$(a,b)$$, then $$y_p=y_{p_1}+y_{p_2}$$ is a solution of

$P_0(x)y''+P_1(x)y'+P_2(x)y=F_1(x)+F_2(x)$

on $$(a,b)$$.

[exer:5.3.40] Suppose $$p$$, $$q$$, and $$f$$ are continuous on $$(a,b)$$. Let $$y_1$$, $$y_2$$, and $$y_p$$ be twice differentiable on $$(a,b)$$, such that $$y=c_1y_1+c_2y_2+y_p$$ is a solution of

$y''+p(x)y'+q(x)y=f$

on $$(a,b)$$ for every choice of the constants $$c_1,c_2$$. Show that $$y_1$$ and $$y_2$$ are solutions of the complementary equation on $$(a,b)$$.